Question

$\int \frac{1}{\sqrt{x}+x\sqrt{x}}dx=$

• A. ${\tan }^{-1}\sqrt{x}+C$
• B. $2\log (\sqrt{x}+1)+C$
• C. $2{\tan }^{-1}\sqrt{x}+C$
• D. $\frac{1}{2}{\tan }^{-1}\sqrt{x}+C$

Answer 1

The correct option is C $2{\tan }^{-1}\sqrt{x}+C$

$\text{Let I}=\int \frac{1}{\sqrt{x}+x\sqrt{x}}dx=\int \frac{1}{\sqrt{x}(1+x)}dx$
$\text{Put}\sqrt{x}=t$
$\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\Rightarrow \frac{1}{2t}dx=dt$
$\therefore \text{I}=\int \frac{2}{1+t^2}dt$
$=2{\tan }^{-1}t+C=2{\tan }^{-1}\sqrt{x}+C$